If kc = 4.0×10−2 for pcl3(g)+cl2(g)⇌pcl5(g) at 520 k , what is the value of kp for this reaction at this temperature?

Here we have to get the of the reaction at 520 K temperature.

The of the reaction is 1.705 atm

We know the relation between and is , where   = The equilibrium constant of the reaction in terms of partial pressure,  = The equilibrium constant of the reaction in terms of concentration and N = number of moles of gaseous products - Number of moles of gaseous reactants.

Now in this reaction, PCl₃ + Cl₂ ⇄ PCl₅

Thus number of moles of gaseous product is 1, and number of moles of gaseous reactants are 2. Thus N = |1 - 2| = 1 mole

The given value of   is 4.0×10⁻²

The molar gas constant, R = 0.082 L. Atm. mol⁻¹. K⁻¹ and temperature, T = 520 K.

On plugging the values in the equation we get,

Or, = 1.705 atm

Thus, the of the reaction is 1.705 atm

it would be d. henry moseley

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